Answer:[tex]\begin{gathered} t=10\log_e47 \\ \\ \approx38.501 \end{gathered}[/tex]
Explanation:
Given the equation:
[tex]e^{0.1t}=47[/tex]
Applying logarithm to both sides:
[tex]\begin{gathered} \log_ee^{0.1t}=\log_e47 \\ \\ 0.1t\log_ee=\log_e47 \\ \\ 0.1t=\log_e47 \\ \\ t=\frac{1}{0.1}\log_e47 \\ \\ t=10\log_e47 \end{gathered}[/tex]
Aproximately, we have:
[tex]t=38.501[/tex]
